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Show that $D_{11} \oplus \mathbb{Z}_3 \not \cong D_3 \oplus \mathbb{Z}_{11}$

The first thing I check is that the size of both groups is the same since

$|D_{11}|=22$ and $|\mathbb{Z}_3|=3$ thus $|D_{11} \oplus \mathbb{Z}_3=22\cdot 3=66$ and likewise $|D_3|=6$ and $|\mathbb{Z}_{11}|=11$ thus $|D_3 \oplus \mathbb{Z}_{11}|=6\cdot 11=66$

Could I evaluate the order of the centers of each group to show they are not the same size thus not isomoprhic? I see that the $|Z(D_{11} \oplus \mathbb{Z}_3)|\not = |Z(D_3 \oplus \mathbb{Z}_{11})|$ since the order of both center of the dihedral groups is only the identity where as the order of the center of $\mathbb{Z}_3$ is not the same size as the order of the center of $\mathbb{Z}_{11}$

HighSchool15
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  • Elements of order 2 – Randall Dec 11 '17 at 18:49
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    Yes, the fact that their centres have different orders shows that they are not isomorphic. – Derek Holt Dec 11 '17 at 18:51
  • There are lots of other approaches as well: For example, In one of these groups there's an element of order 11 that fails to commute with some other element. Is there in the other? It's worth thinking about other possibilities to try to build up your toolbox of skills. – John Hughes Dec 11 '17 at 18:53

1 Answers1

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As noted the center of both groups is different, hence the groups cannot be isomorphic. We have $Z(D_{2n+1})=1$ for all $n\ge 1$ and thus $$ Z(D_{11}\times C_3)=Z(D_{11})\times Z(C_3)=1\times C_3\cong C_3, $$ and $$ Z(D_3\times C_{11})=Z(D_3)\times Z(C_{11})=1\times C_{11}\cong C_{11}. $$

Reference: Center of dihedral group

Dietrich Burde
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